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JEE Advanced 2009Paper 1

Exam: JEE Advanced
Year: 2009
Paper: Paper 1
Subject: Mathematics
Chapter: Circle

JEE Advanced 2009 · Paper 1 · Mathematics · Circle

Tangents drawn from the point P(1,8) to the circle x^2+y^2-6 x-4...

Tangents drawn from the point

P(1,8)

to the circle

x^2+y^2-6 x-4 y-11=0

touch the circle at the points

A

and

B

. The equation of the circumcircle of

\triangle P A B

is

Options

A $x^2+y^2+4 x-6 y+19=0$
B $x^2+y^2-4 x-10 y+19=0$
C $x^2+y^2-2 x+6 y-29=0$
D $x^2+y^2-6 x-4 y+19=0$

Correct answer

B. $x^2+y^2-4 x-10 y+19=0$

Step-by-step solution

For required circle,

P(1,8) and O(3,2)

will be the end point of its diameter.

\begin{aligned} & \therefore(x-1)(x-3)+(y-8)(y-2)=0 \\ & \Rightarrow \quad x²+y²-4 x-10 y+19=0 \end{aligned}

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